Chapter 15: Problem 17

Solve the differential equation $y^{\prime \prime}-a^{2} y=f(t)$, where, $$ f(t)=\left\\{\begin{array}{ll} 0, & t<0 \\ 1, & t>0 \end{array} \quad \text { and } y_{0}=y_{0}^{\prime}=0\right. $$ Hint: Lse the convolution integral as in the example.

Short Answer

Expert verified

$ y(t) = \frac{1}{a^{2}} (1 - \text{ch}(at)) $.

Step by step solution

Understand the Differential Equation

The given differential equation is $y'' - a^{2} y = f(t)$, with initial conditions $y(0) = 0$ and $y'(0) = 0$. The function $f(t)$ is defined piecewise: $f(t) = 0$ for $t < 0$ and $f(t) = 1$ for $t > 0$.

Find the Homogeneous Solution

The homogeneous differential equation is $y'' - a^{2} y = 0$. Solve the characteristic equation $r^{2} - a^{2} = 0$ to find the roots. The roots are $r = a$ and $r = -a$. Thus, the general solution to the homogeneous equation is $y_h(t) = C_1 e^{at} + C_2 e^{-at}$.

Use the Convolution Integral

To find the particular solution, use the convolution integral. The Green's function for the differential equation is given by $G(t - \tau) = \frac{1}{a} \theta(t - \tau) \text{sh}\bigg(a(t - \tau)\bigg)$, where $\theta(t)$ is the Heaviside step function. Because $f(t) = 1$ for $t > 0$, the particular solution is \[ y_p(t) = \int_{0}^{t} G(t - \tau) f(\tau) d\tau = \int_{0}^{t} \frac{1}{a} \theta(t - \tau) \text{sh}\bigg(a(t - \tau)\bigg) d\tau.\]

Simplify the Convolution Integral

Evaluate the integral. The Heaviside step function $\theta(t - \tau)$ ensures that the integral is only considered when $\tau \leq t$. Thus, the integral bounds are from 0 to $t$. We now have \[ y_p(t) = \frac{1}{a} \int_{0}^{t} \text{sh}\bigg(a(t - \tau)\bigg) d\tau.\]

Solve the Integral

Use the fact that the integral of $\text{sh}(at)$ is $\frac{1}{a} \text{ch}(at)$. Therefore, we get \[ y_p(t) = \frac{1}{a} \left[ \frac{1}{a} \text{ch}(a(t - \tau)) \right]_{0}^{t}.\] Evaluating the bounds, \[ y_p(t) = \frac{1}{a^{2}} [\text{ch}(0) - \text{ch}(at)] = \frac{1}{a^{2}} (1 - \text{ch}(at)).\]

Combine Solutions

The total solution is the sum of the homogeneous and particular solutions. Due to the initial conditions $y(0) = 0$ and $y'(0) = 0$, the constants $C_1$ and $C_2$ are zero. Hence, the homogeneous solution $y_h(t)$ is zero, and the total solution is the particular solution: \[ y(t) = \frac{1}{a^{2}} (1 - \text{ch}(at)).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convolution Integral

In the context of differential equations, the convolution integral is a powerful tool used to find particular solutions.
The convolution integral blends a function with the system's response (represented by the Green's function) to produce a solution.
For our problem, it takes the form:
For our problem, it takes the form:
whereas can be thought of as averaging the behavior over multiple shifts of the input function .We integrate over the range of from to ensure that only the part for which the kernel function is defined (considered here in terms of the Heaviside step function ) contributes meaningfully to the overall solution. The Green's function is multiplyed by the forcing function at different instances and integrates over these variations.

Green's Function

A Green's function is essentially an impulse response of a given differential operator.
It helps us understand how the system responds to a delta function (unit impulse). For the differential equation:
We use the Green's function : The Green's function for a linear operator gives us a way to break down complex forcing functions into more manageable parts using the superposition principle.
The Green's function used in this problem is:
for which this The Green's function correcty captures the evaluation of vibratory responses of the system as varies.

Homogeneous Solution

The homogeneous solution of a differential equation is the solution we get when we set the non-Homogenous function (right-hand side of the equation) to zero.

For our problem, the right side was set to zero:
The characteristic equation
The solution gives us the general form:
where and determined through boundary/Initial conditions.
The homogeneous (natural) solutions capture the essential manner by which the system evolves without an external forcing function. In this boundary initial conditions are chosen . These constants happen to be zero in such case and remain left behind by the homogeneous term.

Particular Solution

The term the components natural and applied homogeneous solutions. To capture forced-response of the system to the can implement integral:
These specific problem-by-performing systematic integral transformative to the given forcing function For this problem the particular solution gives.
In our case, the particular solution is:
The general solution is then derived between the summation homogeneous and particular Processes.

Solve the differential equation \(y^{\prime \prime}-a^{2} y=f(t)\), where, $$ f(t)=\left\\{\begin{array}{ll} 0, & t<0 \\ 1, & t>0 \end{array} \quad \text { and } y_{0}=y_{0}^{\prime}=0\right. $$ Hint: Lse the convolution integral as in the example.